A) \[\frac{1}{2}\lambda \]
B) \[\frac{3}{4}\lambda \]
C) \[\frac{2}{3}\lambda \]
D) \[\frac{4}{9}\lambda \]
Correct Answer: D
Solution :
\[hv-\phi =KE\] \[\Rightarrow {{\left( \frac{hc}{\lambda } \right)}_{incident}}=KE+\phi \] \[{{\left( \frac{hc}{\lambda } \right)}_{incident}}\simeq KE\] \[KE=\frac{{{p}^{2}}}{2m}=\frac{hc}{{{\lambda }_{incident}}}=\frac{hc}{\lambda }\] ?(1) \[\Rightarrow \]\[\frac{{{p}^{2}}\times {{(1.5)}^{2}}}{2m}=\frac{hc}{\lambda '}\] ...(2) divide and \[{{(1.5)}^{2}}=\frac{\lambda }{\lambda '}\]\[\Rightarrow \lambda '=\frac{4\lambda }{9}\]You need to login to perform this action.
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